Evaluation of definite Integral Containing Rational functions.. 
Evaluation of $$\int_{-5}^{-2}\left(\frac{x^2-x}{x^3-3x+1}\right)^2dx+\int_{\frac{1}{6}}^{\frac{1}{3}}\left(\frac{x^2-x}{x^3-3x+1}\right)^2dx+\int_{\frac{6}{5}}^{\frac{3}{2}}\left(\frac{x^2-x}{x^3-3x+1}\right)^2dx$$

$\bf{My\; Try::}$ Here Denominator of all rational function is not defined when $x^3-3x+1=0$
So Using Derivative test::
$\displaystyle f(x)=x^3-3x+1\;,$ Then $f'(x)=3x^2-3$ and $f''(x)=6x$
So Foe $\max$ or $\min,$ Put $f'(x)=0\Rightarrow x=\pm 1$
So at $x=-1,$ We get $f''(-1) = -6,$ means $x=-1$ is a point of $\max$ and at $x=1,$
We get $f''(1)=6$ is a point of $\min$
Now drawing a rough graph, We get   roots of $f(x)$ as 
one root lie in $(-2,-1)$ and other root in  $(0,1)$ and third root in$(1,2)$
Now how can i solve after that, Help required, Thanks
 A: The point is to find that the integrand function is continuous at every of these three segments. In other words, you need to proof that the roots of $ f(x) = x^3 - 3x + 1 $ are not in any of the segments at witch you want to integrate. You can do these by analysing the monoton behaviour of these function. For example, $f(-2) < 0$ and you know that the function is monotonically increasing from $-5$ to $-2$, so you are sure about the first integral. 
From $-1 $ to $ 1 $ the function is monotonically decreasing. So, if $ f\big(\frac{1} {6}\big) > 0 $ and  $ f\big(\frac{1} {3}\big) > 0 $ (you have to show that), there is no zero of these function at the segment $\big[\frac{1}{6}, \frac{1}{3} \big]$. 
The same with the third - use the values at the end points and the monotonic behaviour of the function at that segment. (monotonically increasing, $f\big[\frac{6}{5} \big] < 0$, and $f\big[\frac{3}{2} \big] < 0$)
The rest you probably have to do using numerical integration. The usual way with partial fraction decomposition doesn't help. The indefinite integral can't be solved using elementary functions.  
